Convergence Radius: Non-Analyticity Why is a function certainly nonanalytic at some point on the radius of convergence? I mean considering a power series around somewhere and if theres a power series expansion at every point on circle with that radius of convergence would that contradict?

I need that in order to deduce that in a Banach algebra the radius of convergence of power series expansion around infinity agrees with the spectral radius...
 A: Let's call the function $f$, and let $R$ be the radius of convergence of the power series expansion of $f$ about $0$.
If for all $\zeta \in \partial D_R(0)$ there were an analytic continuation $f_\zeta$ of $f$ in a neighbourhood of $\zeta$, we could expand $f_\zeta$ in a power series about $\zeta$, say its radius of convergence is $\rho_\zeta$, and let's assume that $f_\zeta$ is defined on the disk $D_{\rho_\zeta}(\zeta)$ only.
Then, if $D_{\rho_\zeta}(\zeta)\cap D_{\rho_\eta}(\eta) \neq \varnothing$ for $\zeta,\eta\in \partial D_R(0)$, we have $f_\zeta \equiv f_\eta$ on $D_{\rho_\zeta}(\zeta)\cap D_{\rho_\eta}(\eta) \cap D_R(0)$, and by the identity theorem on all of $D_{\rho_\zeta}(\zeta)\cap D_{\rho_\eta}(\eta)$, since that is connected. But then we would have an analytic continuation $\tilde{f}$ of $f$ to a disk $D_{R+\varepsilon}(0)$ furnished by the local continuations $f_\zeta$, and then Cauchy's integral formula
$$\tilde{f}(z) = \frac{1}{2\pi i} \int_{\lvert \zeta\rvert = r} \frac{\tilde{f}(\zeta)}{\zeta-z}\,d\zeta = \sum_{n=0}^\infty \left(\frac{1}{2\pi i}\int_{\lvert\zeta\rvert = r} \frac{\tilde{f}(\zeta)}{\zeta^{n+1}}\,d\zeta\right)z^n$$
shows that the Taylor series of $f$ converges in the disk $D_r(0)$ for all $r < R+\varepsilon$, and hence the radius of convergence of the Taylor series must be $\geqslant R+\varepsilon$, contradiction the assumption that the radius of convergence is $R$.
